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Mathematics 17 Online
OpenStudy (anonymous):

Another Calculus question comin' atcha! Screenshot in the comments.

OpenStudy (anonymous):

OpenStudy (anonymous):

Any ideas from your last question?

OpenStudy (anonymous):

Hint: fundamental theorem of calc \[\frac{d}{dx}\int_c^{g(x)}f(t)~dt=f(g(x)) \times g'(x)\]

OpenStudy (anonymous):

Are they related at all? would it be 2x(x^2^2)?

OpenStudy (anonymous):

Not quite, with \(f(t)=t^t\), this makes \(f(g(x))=\big[g(x)\big]^{g(x)}\).

OpenStudy (anonymous):

In this case, \(x^2\) is being raised to the power of \(x^2\), so you have \[\big(x^2\big)^{x^2}=x^{2x^2}\]

OpenStudy (anonymous):

Ok, it seems I got that part, 2x(x^2)x^2, so now I just plug in 3 for x?

OpenStudy (anonymous):

Yup! I got the answer right. Thanks a bunch!

OpenStudy (anonymous):

Yes, provided you have the right exponents and such

OpenStudy (anonymous):

yw

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